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{\bf xxxx}$^{a}${\footnote { email: {\sf xxxx}}} and {\bf
yyyy}$^{b,c}$ {\footnote { email: {\sf yyyy}}}

\vspace{0.5cm}
$^{a}${\it Department of Physics , Faculty of Sciences, University Ibnou Zohr,\\
 Agadir ,
Morocco}\\[1em]

$^{b}${\it LPHE-Modeling and Simulation, Faculty  of Sciences,
University
Mohammed V,\\ Rabat, Morocco}\\[1em]

$^{c}${\it Centre of Physics and Mathematics,
CPM, CNESTEN,\\ Rabat, Morocco}\\[1em]


\vspace{3cm} {\bf Abstract}
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\medskip

We study the evolution of geometric quantum discord (GMQD) of a two
qubits system coupled with two independent bosonic reservoirs. We
consider sub-ohmic, ohmic and super-ohmic. A special attention is
devoled to Dicke states and their superpositions.


\newpage
\section{Introduction}




Multipartite entanglement is one of the important areas in the field
of quantum information that has many applications including quantum
secret sharing. In this paper, we focus on the Dicke
states~\cite{PhysRev.93.99}, which are useful building blocks in
realizing multipartite entanglement. The $n$-qubit weight $w$ Dicke
state, $|D^n_w\rangle$, is the equal superposition of all $n$-qubit
states of weight $w$. We refer to~\cite{dickold,d2,d3,phyb,d1,d4,d5}
and the references therein for detailed discussion.

After the invention of quantum information, many experimental setups
have been proposed and tested to verify some theoretical properties.
Most of experiments have been focused on the test of multipartite
entanglement such as EPR, GHZ, and W states. Since the result of
experimental tests depends on the steps for preparing, processing,
and measuring, all steps should be refined as much as possible.
Among them, the first priority is to prepare the target state with
very high fidelity and with efficiency. In this work, therefore, we
also focus on the efficient way to prepare certain multipartite
quantum state.

In line of GHZ and W states, we have the Dicke state,
$|D^n_w\rangle$, which an equal superposition state of all $n$-qubit
states of weight $w$. Actually, Dicke state is more general state
than GHZ and W states since W state is $|D^n_1\rangle$ and GHZ state
is the superposition of $|D^n_0\rangle$ and $|D^n_n\rangle$.
Therefore, the preparation method for Dicke state can be utilized
for other general case as well. At the same time, similar to the
above reason, Dicke state can be utilized for many applications such
as secret sharing \cite{d5} and quantum networking \cite{dd1}.
Related to this, some previous works have been done that focussed on
the experimental ways to prepare six-qubit Dicke state \cite{d5,d4}
with fidelity $0.654\pm0.024$ and $0.56\pm0.02$, respectively.

While the main focus from the viewpoint of experimental physics is
to actually provide the implementation of specific Dicke states, our
focus is from theoretical algorithmic angle and the only result
presented in this direction appeared in~\cite{dickold}. In this
work, we show how one can efficiently construct Dicke states by
using the combinatorial properties of symmetric Boolean functions,
two well-known quantum algorithms, and the generalized parity
measurement. By efficient, we mean that the resource requirements in
terms of quantum circuits and number of execution steps is poly$(n)$
to obtain $|D^n_w\rangle$.

Let us consider $n$-qubit states in the computational basis $\{0,
1\}^n$ that can be written in the form $\sum_{x \in \{0, 1\}^n} a_x
|x\rangle$, where $\sum_{x \in \{0, 1\}^n} |a_x|^2 = 1$. Thus, $x$
can also be interpreted as a binary string and the number of $1$'s
in the string is called the (Hamming) weight of $x$ and denoted as
$wt(x)$. Based on this an arbitrary Dicke state can be expressed as
follows:
$$|D^n_w\rangle =\sum\limits_{x\in\{0,1\}^n,wt(x)=w}\frac{1}{\sqrt{\binom{n}{w}}}|x\rangle.$$
Let us also define a symmetric $n$-qubit state as
$$|S^n\rangle=\sum_{x\in\{0,1\}^n}a_{wt(x)}|x\rangle, \mbox{ where }
\sum_{i=0}^n\binom{n}{i}|a_i|^2=1.$$

First, we show how one can prepare a symmetric $n$-qubit state with
the property that $\binom{n}{w}|a_w|^2$ is
$\Omega(\frac{1}{\sqrt{n}})$ by using Deutsch-Jozsa
algorithm~\cite{qDJ92}. This requires certain novel combinatorial
observations related to symmetric Boolean functions. Then the
quantum state out of Deutsch-Jozsa algorithm is measured using the
parity measurement technique~\cite{d2} to obtain $|D^n_w\rangle$
with a probability $\Omega(\frac{1}{\sqrt{n}})$. Thus, $O(\sqrt{n})$
runs are sufficient to obtain the required Dicke state. Note that a
direct approach to construct a symmetric state has been presented
in~\cite{dickold} using biased Hadamard transform. While the order
of probability to obtain Dicke state by ours and that
of~\cite{dickold} are the same, enumeration results show that the
exact probability values are better in our case than that
of~\cite{dickold}.

Further, motivated by the idea in~\cite{dickold}, we improve our
algorithm further with a modified Deutsch-Jozsa operator that
involves the biased Hadamard transform. Since biased Hadamard
transform also helps to generate the target symmetric state, the
overall probability to obtain the Dicke state increases.

Finally, we can also apply the Grover operator~\cite{qGR96} before
the measurement. Since Grover algorithm amplifies the amplitude of
target symmetric state, this helps to reduce the necessary number of
steps into $O(\sqrt[4]{n})$.




\section{Majorana Representation} \label{Sec: Appendix}

%\noindent {\bf{Majorana representation}}

The permutation symmetric subspace of $n$ qubits is spanned by the
Dicke states
\begin{align}
|S(n,k)\rangle := \frac{1}{\sqrt{{n\choose
k}}}(\sum_{PERM}|\underbrace{00...0}_{n-k}\underbrace{11..1}_{k}\rangle),
\end{align}
which can be understood as the symmetric states with $k$
excitations. Thus any permutation symmetric state can be written as
\begin{align}
|\psi\rangle = \sum a_k |S(n,k)\rangle.
\end{align}
Alternatively all symmetric states can be written in the Majorana
representation \cite{Majorana}
\begin{eqnarray} \label{Eqn: App Symm state}
|\psi\rangle =\frac{e^{i\alpha}}{\sqrt{K}} \sum_{PERM}
|\eta_1\rangle |\eta_2\rangle...|\eta_n\rangle,
\end{eqnarray}
where the sum is over all permutations and $K$ is a normalisation
constant.

To find the Majorana representation (\ref{Eqn: App Symm state}) we
consider the overlap with product state $|\phi\rangle^{\otimes
n}$,$|\phi\rangle=\cos\left(\frac{\theta}{2}\right)|0\rangle +e^{i
\varphi} \sin\left(\frac{\theta}{2}\right)|1\rangle$. It is clear by
comparison to equation (\ref{Eqn: App Symm state}) that
$|\phi\rangle$ orthoganol to the MP $|\eta_i\rangle$ will give zero
overlap. This is exactly how we find the MPs. For simplicity we take
a multiple of the overlap, sometimes called the {\it characteristic
polynomial}, {\it Majorana polynomial}, {\it amplitude function} or
{\it coherent state decomposition}
\begin{align} \label{Eqn: Maj functional}
f(\psi):= \cos^{-n}\left(\frac{\theta}{2}\right) \langle
\phi|^{\otimes n}|\psi\rangle = \sum_{k=0}^n \sqrt{{n \choose k}}
a_k \alpha^k,
\end{align}
which is a complex polynomial in
$\alpha:=e^{-i\varphi}\tan\left(\frac{\theta}{2}\right)$. By the
fundamental theorem of algebra this has unique zeros up to
multiplication by some complex. Hence the zeros $\alpha_j =
e^{-i\varphi_j}\tan\left(\frac{\theta_j}{2}\right)$ define the state
$|\psi\rangle$ up to a global phase. The corresponding MPs are at
position $\theta_j'=\theta_j+\pi$,~$\varphi_j'=\varphi_j+\pi$.

Note that we can understand the state $|\phi\rangle^{\otimes n}$ as
a kind of generalized coherent state \cite{Perelomov,Klauder},
defined by the action of a group on some chosen fiducial state (so
that certain properties apply such as overcompleteness). For our
case the group is SU(2) as represented by $U^{\otimes n}$, with $U$
a rotation through $\theta, \varphi$ and the fiducial state
$|0\rangle^{\otimes n}$, that is
\begin{eqnarray}
|\phi\rangle^{\otimes n}= U^{\otimes n} |0\rangle^{\otimes n}.
\end{eqnarray}
When viewing the symmetric subspace as one spin $S=n/2$ system,
these are equivalent to spin coherent states
\cite{Arecchi72,Radcliffe71}. In this sense the Majorana
representation is a kind of condensed coherent state representation
of states (since it is only concerned with the zeros of the coherent
state decomposition (\ref{Eqn: Maj functional})).




The Majorana representation then allows us to identify symmetries to
show equality of the entanglement measures for many new sets of
states. The complete set of all the possible subgroups of SO(3) are
the continuous groups, orthogonal $O(2)$ and special orthogonal
$SO(2)$, and discrete groups Cyclic $C_m$, Dihedral $D_m$,
Tetrahedral $T$, Octehedral $O$ and Isocahedral $Y$. One can then
systematically go through all of these groups to find these special
states, as done in \cite{Makela07} in the context of inert states.
For the subgroup of arbitrary rotations about a fixed axis SO(2), we
see that states with MPs only at either pole of the rotation axis
satisfy our condition. If the rotations are around the $Z$-axis,
these are the states
\begin{align}
|S(n,k)\rangle := \frac{1}{\sqrt{{n\choose
k}}}(\sum_{PERM}|\underbrace{00...0}_{n-k}\underbrace{11..1}_{k}\rangle),
\nonumber
 \end{align}
also known as Dicke states, and we can see here pictorially the
proof of equivalence for these states reported in \cite{Hayashi08}.
Note that for even $n$ and $k=n/2$, these states also satisfy our
condition for the group $O(2)$ (arbitrary rotations around the
$Z$-axis, and a flip on some axis in the $X-Y$ plane). In such cases
we associate the state with the smallest subgroup. The cyclic group
$C_n$ has no truly invariant states - since if all points are moved
together up and down the axis of rotation the symmetry is not lost.
The dihedral group $D_m$ (consisting of rotations through $2\pi/m$
and a flip on the axis of rotation) has $m$ totally invariant states
for each value $m$ (see Fig.~\ref{FIG: dihedral}). $T$, $O$ and $Y$
only have truly invariant states for certain $n$. For the
tetrahedral group $T$ truly invariant states are made up of
tetrahedrons, their antipode tetrahedrons, and octagons with at most
$2$ MPs on any tetrahedron point and $3$ MPs at any octahedron
point, so that there are only truly invariant states for $n\leq 34$.
For the octahhedral group $O$ truly invariant states have MPs at the
points of the cube and the octahedron with at most $3$ and $2$ MPs
at each respectively, so that they only exist for $n\leq34$. All
truly invariant states of the Isocahedral group $Y$ are made up of
combinations of isocahedrons (with $12$ vertices) and dodecahedrons
(with $20$ vertices) with at most $3$ and $2$ MPs at each
respectively, hence they exist only for $n\leq88$.  For four qubits
there are four entangled states satisfying the condition,
$|T\rangle=1/\sqrt{3}|S(4,0)\rangle+\sqrt{2/3}|S(4,3)\rangle$,
$|GHZ_4\rangle$, $|S(4,2)\rangle$ and $|W_4\rangle=|S(4,1)\rangle$
as shown in Fig.~\ref{FIG: test}.




%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Symmetric multi-qubit systems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


The multi-qubit symmetric states were shown relevant  for different
purposes  in quantum information science~\cite{solano,
mixed,usa1,usa2,markham1,markham2,gebastin,markham3}. In this paper,
we shall mainly focus on an ensemble of $n$ spin-$1/2$ prepared in
even and
 odd spin coherent states.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Spin coherent as  symmetric multi-qubit systems}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

We consider $n$ identical qubits. Each qubit lives in a
2-dimensional Hilbert space
$\mathcal{H}=\mathrm{span}\{\vert{0}\rangle,\vert{1}\rangle\}$. The
Hilbert space of the $n$-qubit system is given  by $n$ tensored
copies of $\mathcal{H}$
$$\mathcal{H}_n := \mathcal{H}^{\otimes n}.$$
Among the multi-partite states in $\mathcal{H}_n$, multi-qubit
states obeying exchange symmetry are of special interest from
experimental as well as mathematical point of views. An arbitrary
symmetric $n$-qubit state is commonly represented in either
Majorana~\cite{Maj32} or  Dicke ~\cite{Dic54} representation. Any
multi-qubit state, invariant under the exchange symmetry, is
specified in the Majorana description by the state (up to a
normalization factor)
\begin{equation}\label{psisym}
|\psi_s\rangle = \frac{1}{n!} \sum_{\sigma \in {\cal S}_n} \vert
{\eta_{\sigma(1)},\ldots,\eta_{\sigma(n)}}\rangle,
\end{equation}
where each single qubit state is $|\eta_i\rangle \equiv (1 + \eta_i
\bar\eta_i)^{-\frac{1}{2}}(|0\rangle + \eta_i |1\rangle)~
(i=1,\ldots, n)$
%$|\eta_i\rangle \equiv \alpha_i |0\rangle + \beta_i |1\rangle$ ($i=1,\ldots, n$)
 and
the sum is over the elements of the permutation group ${\cal S}_n$
of $n$ objects. In Equation (\ref{psisym}), the vector
$\vert{\eta_{\sigma(1)},\ldots,\eta_{\sigma(n)}}\rangle$ stands for
the tensor product
$\vert{\eta_{\sigma(1)}}\rangle\otimes\ldots\otimes\vert{\eta_{\sigma(n)}}\rangle$.
The totally symmetric $n$-qubit states can be also formulated  in
Dike representation. The symmetric Dicke states with $k$ excitations
are defined by ~\cite{Dic54}
%\begin{equation}
%\label{Dicke}
%\vert {D_n(k)} \rangle = \frac{1}{\sqrt{C_n^k}} \sum_{\sigma} \vert{\underbrace{0\ldots 0}_{n-k}\underbrace{1\ldots 1}_{k}}\rangle,
%\end{equation}
\begin{equation}
\label{Dicke} \vert n,k \rangle = \sqrt{\frac{k!(n-k)!}{n!}}
\sum_{\sigma \in {\cal S}_n} \vert{\underbrace{0,\ldots,
0}_{n-k},\underbrace{1,\ldots, 1}_{k}}\rangle,
\end{equation}
which generate an orthonormal basis of the  symmetric Hilbert
subspace of dimension $(n+1)$. Therefore, permutation invariance,
in symmetric multi-qubit states, implies a  restriction to $n+1$
dimensional subspace from the entire  $2^n$ dimensional Hilbert
space. The Dicke states (\ref{Dicke}) constitute a special subset of
the symmetric multi-qubit states (\ref{psisym}) corresponding to the
situation where the first $k$-qubit are such that $\eta_i = 0$ for
$i = 0,1, \ldots k$ and the remaining qubits are in the states
$\vert \eta_i = 1 \rangle $ with $i = k+1, \ldots , n$.  Any
symmetric state  $|\psi_s \rangle$ (\ref{psisym}) can be expanded in
terms of Dicke states (\ref{Dicke}) as follows
\begin{equation}
\label{psisymD} |\psi_s\rangle = \frac{1}{n!} \sum_{k=0}^n c_k ~|n,k
\rangle,
\end{equation}
where  the $c_k$ ($k=0,\ldots,n$) stand for the complex expansion
coefficients. In particular, when the qubit are all identical
($\eta_i = \eta$ for all qubits), it is simply verified that the
coefficients $c_k$ are given by
\begin{equation}
\label{ck} c_k = n!  \sqrt{\frac{n!}{k!(n-k)!}} \frac{\eta^{k}}{(1 +
\eta\bar\eta)^{\frac{n}{2}}}
\end{equation}
and the symmetric multi-qubit states  (\ref{psisym}) write
\begin{equation}\label{psisym-cs}
|\psi_s\rangle :=  \vert n, \eta \rangle = (1 +
\eta\bar\eta)^{-\frac{n}{2}} \sum_{k=0}^n \sqrt{\frac{n!}{k!(n-k)!}}
\eta^{k}  ~|n,k \rangle,
\end{equation}
which are exactly the  $j=\frac{n}{2}$-spin coherent states (for
more details see for instance \cite{daoud2}). In particular, the
state $\vert n, \eta \rangle $ can be identified for $n=1$ with
spin-$\frac{1}{2}$ coherent state with $|0\rangle \equiv
|\frac{1}{2}, -\frac{1}{2}\rangle$ and $|1\rangle \equiv
|\frac{1}{2} ,+ \frac{1}{2}\rangle$)





%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Multi-qubit "Shro\"odinger cat" states}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

The prototypical multi-qubit "Shr\"odinger cat" states, we consider
in this work, are defined  as a balanced superpositions of the
$n$-qubit states $\vert n, \eta \rangle$ and $\vert n, -\eta
\rangle$ given by (\ref{psisym-cs}). They write
\begin{equation}\label{ncs}
 \vert \eta, n , m \rangle  =  {\cal N}( \vert n, \eta \rangle + e^{im\pi} \vert n, - \eta
 \rangle)
\end{equation}
where
$$ \vert n, \pm \eta \rangle = \vert \pm \eta \rangle \otimes \vert \pm \eta \rangle \cdots \otimes \vert \pm \eta \rangle, $$
and the integer $m \in \mathbb{Z}$ takes the values $m = 0 ~({\rm
mod}~2)$ and $m = 1~ ({\rm mod}~2)$. The normalization factor ${\cal
N}$ is
$$ {\cal N}= \big[ 2 + 2 p^{n} \cos m \pi\big]^{-1/2}$$
where  $p$ denotes the overlap between the states $\vert \eta
\rangle$ and $\vert  -\eta \rangle$. It is given by
\begin{equation}\label{overlap}
 p = \langle \eta \vert  - \eta \rangle = \frac{1 - \bar\eta \eta}{1 + \bar\eta \eta}.
\end{equation}
 Experimental creation of cat states
comprising multiple particles was reported in the literature
\cite{Wineland,Haroche}. Due to their experimental implementation,
"Shr\"odinger cat" states are expected to be an useful resource for
quantum computing as well as quantum communications. Also, in view
of their mathematical elegance, multi-qubit states obeying exchange
symmetry offer drastic simplification  in investigating various
aspects of quantum correlations in particular the geometric measure
of quantum discord as we shall discuss in the present work.
Furthermore,  the multi-qubit symmetric states (\ref{ncs}) include
Greenberger-Horne-Zeilinger(${\rm GHZ}$)~\cite{ghz}, ${\rm W}$
\cite{Dur00} and Dicke states~\cite{Dic54}. The multi-qubits states
$ \vert n, \eta , 0 \rangle$ ($m=0 ~ {\rm mod}~ 2$) and $ \vert n,
\eta , 1 \rangle$ ($m=1 ~ {\rm mod}~ 2$) behave like a multipartite
state of Greenberger-Horne-Zeilinger (${\rm GHZ}$) type \cite{ghz}
in the limiting case $p \rightarrow 0 $. Indeed,
 the  states $|\eta \rangle $ and $|
-\eta \rangle $ approach orthogonality and an orthogonal basis can
be defined such that $\vert {\bf 0}\rangle\equiv \vert \eta \rangle$
and $\vert{\bf 1}\rangle \equiv \vert  -\eta \rangle$. Thus, the
state $\vert n , \eta, m \rangle$ becomes of ${\rm GHZ}$-type:
\begin{equation}
\vert \eta , n, m \rangle \sim \vert {\rm GHZ}\rangle_{n} = \frac
1{\sqrt{2}}(\vert {\bf 0}\rangle \otimes |{\bf 0}\rangle \otimes
        \cdots \otimes\vert {\bf 0}\rangle
    +e^{i m \pi}\vert {\bf 1}\rangle \otimes
    \vert {\bf 1}\rangle \otimes \cdots \otimes
\vert {\bf 1}\rangle).\label{GHZ}
\end{equation}
Also, in the special situation where the overlap $p$ tends to unity
($p \rightarrow 1$ or $ \eta \rightarrow 0$ ), the state $\vert \eta
, n, m = 0 ~({\rm mod}~ 2) \rangle$ (\ref{ncs}) reduces to ground
state of a collection of $n$ qubits
\begin{equation}
\vert 0,  n , 0 ~({\rm mod}~ 2) \rangle \sim  \vert 0 \rangle
\otimes\vert 0 \rangle \otimes \cdots \otimes \vert 0 \rangle,
\end{equation}
and  it is simple to check that the state $\vert \eta , 0 , 1 ~({\rm
mod}~ 2) \rangle$  becomes a multipartite state of $W$
type~\cite{Dur00}
\begin{equation}
\vert 0,  n ,  1 ~({\rm mod}~ 2) \rangle \sim \vert\text{\rm
W}\rangle_{n}
    = \frac{1}{\sqrt{n}}(\vert 1 \rangle \otimes\vert 0\rangle \otimes \cdots\otimes
       \vert 0 \rangle  +\vert 0 \rangle \otimes\vert 1 \rangle \otimes\ldots\otimes \vert 0 \rangle
      +\cdots
   + \vert  0 \rangle \otimes\vert 0 \rangle  \otimes \cdots\otimes \vert 1 \rangle)~.
\label{Wstate}
\end{equation}
It is clear that the Shr\"odinger cat  states  $\vert \eta, n , m =
0 ~({\rm mod}~2) \rangle$  include the  ${\rm GHZ}_{n}$ states $(p
\rightarrow 0)$. In other hand, the states
 $\vert \eta, n , m = 1 ~({\rm mod}~2) \rangle$, constitute an interpolation between
two special classes of multi-qubits states: $\vert\text{\rm
GHZ}\rangle_{n}$ type corresponding to  $p \rightarrow 0$ and states
of $\vert\text{\rm W}\rangle_{n}$ type obtained in the special case
where $p \rightarrow 1$.





\section{Conclusion}
In this paper, in order to investigate the pairwise quantum
correlations, we have studied the dynamics of GMQD The model under
consideration consists of two qubits coupled with two independent
bosonic reservoirs described by Ohmic-like spectral densities. The
Hamiltonian of the model is described as Eq (\ref{z}), and the
two-qubit system is initially in the X states decoupled from the
environments, described by Eq (\ref{v}). We examine the evolution of
GMQD for the three types of reservoirs, sub-ohmic, ohmic, and
super-ohmic particularly the Dicke states and their superpositions.
$D_G$ is a monotonic decreasing function of time t or a poiecewise
monotonic decreasing function with one turning point before becoming
frozen phenomenon.

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\end{document}




We show how stationary entanglement between an optical cavity field
mode and a macroscopic vibrating mirror can be generated by means of
radiation pressure. We also show how the generated optomechanical
entanglement can be quantified and we suggest an experimental
readout-scheme to fully characterize the entangled state.
Surprisingly, such optomechanical entanglement is shown to persist
for environment temperatures above 20K using state-of-the-art
experimental parameters.
\end{abstract}

\pacs{03.67.Mn, 42.50.Lc, 05.40.Jc}

\maketitle


Entanglement, ``\textit{the} characteristic trait of quantum
mechanics''~\cite{Schrodinger35}, has raised widespread interest in
different branches of physics. It provides insight into the
fundamental structure of physical reality~\cite{Bell64} and it has
become a basic resource for many quantum information processing
schemes~\cite{entmicro1}. So far entanglement has been
experimentally prepared and manipulated using microscopic quantum
systems such as photons, atoms and ions~\cite{entmicro1,entmicro2}.
Nothing in the principles of quantum mechanics prevents macroscopic
systems to attain entanglement. However, the answer to the question
as to what extent entanglement should hold when going towards
``classical'' systems is yet unknown~\cite{Leggett}. Therefore it is
of crucial importance to investigate the possibilities to obtain
entangled states of macroscopic systems~\cite{Armour02} and to study
the robustness of entanglement against temperature~\cite{Vedral}.
Experiments in this direction include single-particle interference
of macro-molecules~\cite{Hackermueller}, the demonstration of
entanglement between collective spins of atomic
ensembles~\cite{Polzik}, and of entanglement in Josephson-junction
qubits~\cite{Berkley}. Mechanical oscillators are of particular
interest since they resemble a prototype of ``classical'' systems.
Thanks to the fast-developing field of microfabrication, micro- or
nano-mechanical oscillators can now be prepared and controlled to a
very high precision \cite{Roukes03}. In addition, several
theoretical proposals exist that suggest how to reach the quantum
regime for such systems \cite{Pinard0}. Experimentally, quantum
limited measurements have been developed that could allow ground
state detection~\cite{LaHaye}. However, quantum effects in
mechanical oscillators have not been demonstrated to date.

Optomechanical coupling via radiation pressure~\cite{Braginsky} is a
promising approach to prepare and manipulate quantum states of
mechanical oscillators. Proposals range from the quantum state
transfer from light to a mechanical oscillator to entangling two
such oscillators \cite{Braunstein,Marshall03,pinard2,Tombesi}. In
this paper we propose an experimental scheme to create and probe
optomechanical entanglement between a light field and a mechanical
oscillator. This is achieved using a bright laser field that
resonates inside a cavity and couples to the position and momentum
of a moving (micro)mirror.







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-0.44 L cp clip np
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\begin{document}
\title{Efficient quantum algorithm to construct arbitrary Dicke states}
\author{Kaushik Chakraborty$^1$, Byung-Soo Choi$^2$, Arpita Maitra$^1$ and Subhamoy Maitra$^1$}
\affiliation{$^1$Applied Statistics Unit, Indian Statistical Institute, 203 B T Road, Kolkata 700 108, India,\\
email: kaushik.chakraborty9@gmail.com, arpita76b@rediffmail.com, subho@isical.ac.in\\
$^2$Department of Electrical and Computer Engineering, Duke
University, Durham, NC 27708, USA, email: bschoi3@gmail.com}
%\date{}
\begin{abstract}
In this paper, we study efficient algorithms towards the

construction of any arbitrary Dicke state. Our contribution is to
use proper symmetric Boolean functions that involve manipulations
with Krawtchouk polynomials. Deutsch-Jozsa algorithm, Grover
algorithm and the parity measurement technique are stitched together
to devise the complete algorithm. Further, motivated by the work of
Childs et al (2002), we explore how one can plug the biased Hadamard
transformation in our strategy. Our work compares fairly with the
results of Childs et al (2002).
\end{abstract}
\pacs{03.65.Wj, 03.67.Ac, 03.67.Lx} \maketitle \noindent{\bf
Keywords:} Biased Hadamard Transform, Deutsch-Jozsa Algorithm, Dicke
State, Grover Algorithm, Krawtchouk Polynomial, Parity Measurement,
Symmetric Boolean Functions.

\section{Introduction}
Multipartite entanglement is one of the important areas in the field
of quantum information that has many applications including quantum
secret sharing. In this paper, we focus on the Dicke
states~\cite{PhysRev.93.99}, which are useful building blocks in
realizing multipartite entanglement. The $n$-qubit weight $w$ Dicke
state, $|D^n_w\rangle$, is the equal superposition of all $n$-qubit
states of weight $w$. We refer to~\cite{dickold,d2,d3,phyb,d1,d4,d5}
and the references therein for detailed discussion.

After the invention of quantum information, many experimental setups
have been proposed and tested to verify some theoretical properties.
Most of experiments have been focused on the test of multipartite
entanglement such as EPR, GHZ, and W states. Since the result of
experimental tests depends on the steps for preparing, processing,
and measuring, all steps should be refined as much as possible.
Among them, the first priority is to prepare the target state with
very high fidelity and with efficiency. In this work, therefore, we
also focus on the efficient way to prepare certain multipartite
quantum state.

In line of GHZ and W states, we have the Dicke state,
$|D^n_w\rangle$, which an equal superposition state of all $n$-qubit
states of weight $w$. Actually, Dicke state is more general state
than GHZ and W states since W state is $|D^n_1\rangle$ and GHZ state
is the superposition of $|D^n_0\rangle$ and $|D^n_n\rangle$.
Therefore, the preparation method for Dicke state can be utilized
for other general case as well. At the same time, similar to the
above reason, Dicke state can be utilized for many applications such
as secret sharing \cite{d5} and quantum networking \cite{dd1}.
Related to this, some previous works have been done that focussed on
the experimental ways to prepare six-qubit Dicke state \cite{d5,d4}
with fidelity $0.654\pm0.024$ and $0.56\pm0.02$, respectively.

While the main focus from the viewpoint of experimental physics is
to actually provide the implementation of specific Dicke states, our
focus is from theoretical algorithmic angle and the only result
presented in this direction appeared in~\cite{dickold}. In this
work, we show how one can efficiently construct Dicke states by
using the combinatorial properties of symmetric Boolean functions,
two well-known quantum algorithms, and the generalized parity
measurement. By efficient, we mean that the resource requirements in
terms of quantum circuits and number of execution steps is poly$(n)$
to obtain $|D^n_w\rangle$.

Let us consider $n$-qubit states in the computational basis $\{0,
1\}^n$ that can be written in the form $\sum_{x \in \{0, 1\}^n} a_x
|x\rangle$, where $\sum_{x \in \{0, 1\}^n} |a_x|^2 = 1$. Thus, $x$
can also be interpreted as a binary string and the number of $1$'s
in the string is called the (Hamming) weight of $x$ and denoted as
$wt(x)$. Based on this an arbitrary Dicke state can be expressed as
follows:
$$|D^n_w\rangle =\sum\limits_{x\in\{0,1\}^n,wt(x)=w}\frac{1}{\sqrt{\binom{n}{w}}}|x\rangle.$$
Let us also define a symmetric $n$-qubit state as
$$|S^n\rangle=\sum_{x\in\{0,1\}^n}a_{wt(x)}|x\rangle, \mbox{ where }
\sum_{i=0}^n\binom{n}{i}|a_i|^2=1.$$

First, we show how one can prepare a symmetric $n$-qubit state with
the property that $\binom{n}{w}|a_w|^2$ is
$\Omega(\frac{1}{\sqrt{n}})$ by using Deutsch-Jozsa
algorithm~\cite{qDJ92}. This requires certain novel combinatorial
observations related to symmetric Boolean functions. Then the
quantum state out of Deutsch-Jozsa algorithm is measured using the
parity measurement technique~\cite{d2} to obtain $|D^n_w\rangle$
with a probability $\Omega(\frac{1}{\sqrt{n}})$. Thus, $O(\sqrt{n})$
runs are sufficient to obtain the required Dicke state. Note that a
direct approach to construct a symmetric state has been presented
in~\cite{dickold} using biased Hadamard transform. While the order
of probability to obtain Dicke state by ours and that
of~\cite{dickold} are the same, enumeration results show that the
exact probability values are better in our case than that
of~\cite{dickold}.

Further, motivated by the idea in~\cite{dickold}, we improve our
algorithm further with a modified Deutsch-Jozsa operator that
involves the biased Hadamard transform. Since biased Hadamard
transform also helps to generate the target symmetric state, the
overall probability to obtain the Dicke state increases.

Finally, we can also apply the Grover operator~\cite{qGR96} before
the measurement. Since Grover algorithm amplifies the amplitude of
target symmetric state, this helps to reduce the necessary number of
steps into $O(\sqrt[4]{n})$.

\section{Properties of Symmetric Boolean Functions}
\label{subbool}

\subsection{Walsh Spectrum of Symmetric Boolean Functions}

A Boolean function on $n$ variables may be viewed as a mapping from
$\{0,1\}^n$ into $\{0,1\}$. Let us denote the addition operator over
$GF(2)$ by $\oplus$. Let ${x}=(x_1,\ldots,x_n)$ and
${\omega}=(\omega_1,\ldots,\omega_n)$ both belong to $\{0,1\}^n$ and
the inner product
$${x}\cdot{\omega}=x_1\omega_1\oplus\cdots\oplus x_n\omega_n.$$
Let $f({x})$ be a Boolean function on $n$ variables. Then the {\em
Walsh transform} of $f({x})$ is a real valued function over
$\{0,1\}^n$ which is defined as
$$W_f({\omega})=\sum_{{x}\in\{0,1\}^n}(-1)^{f({x})\oplus{x}\cdot{\omega}}.$$

An $n$-variable Boolean function $f$ is called Symmetric if
$f(x)=f(y)$ for all $x, y \in\{0,1\}^n$ such that $wt(x)=wt(y)$.
Henceforth, we will denote the set of $n$-variable symmetric Boolean
functions as ${\cal SB}_n$.

In the truth table of $f \in {\cal SB}_n$, it is enough to provide
outputs corresponding to different weights of elements of $\{0,
1\}^n$ only. So an $n$-variable symmetric function can be expressed
by an $(n+1)$ length bit string as
$$re_f=[f_0,f_1,\ldots,f_n],$$
where $f_i$ is the output at the inputs of weight $i$ and $re_f$ is
referred to as the simplified value vector. When $f\in{\cal SB}_n$,
one may note that $W_f(x)=W_f(y)$ for all $x,y\in\{0,1\}^n$ such
that $wt(x) = wt(y)$. Therefore, the Walsh spectrum of $f$ can be
represented by an $(n+1)$ length integer string
$$rw_f=[rw_f(0),rw_f(1),\ldots,rw_f(n)],$$
where $rw_f(i)$ represents the Walsh spectrum value at the inputs of
weight $i$.

\subsection{Relation between Walsh Spectrum and Krawtchouk polynomials}
We now relate the Walsh spectrum of the symmetric
functions~\cite{kws} with Krawtchouk polynomials~\cite{IK96,FJ77}.
Krawtchouk polynomial of degree $i$ is given by
$$K_i(\eta,n)=\sum_{j=0}^i(-1)^j\binom{\eta}{j}\binom{n-\eta}{i-j}.$$
From~\cite{kws}, we get that if $wt(\omega)=k$, then
$$W_f(\omega)=\sum_{i=0}^n(-1)^{f_i}K_i(k, n).$$ The $(n+1)\times(n+1)$ matrix
which has $K_i(k,n)$ as the $(i,k)$-th element is known as the
Krawtchouk matrix~\cite{krawten,fein}.

For example, let us present the Krawtchouk matrix for $n = 5$ and
$6$ as follows:
\begin{center}
$\left[
\begin{array}{rrrrrr}
 1   &   1   &   1   &   1   &   1   &    1 \\
 5   &   3   &   1   &  -1   &  -3   &   -5 \\
10   &   2   &  -2   &  -2   &   2   &   10 \\
10   &  -2   &  -2   &   2   &   2   &  -10 \\
 5   &  -3   &   1   &   1   &  -3   &    5 \\
 1   &  -1   &   1   &  -1   &   1   &   -1
\end{array}
\right],$ $\left[
\begin{array}{rrrrrrr}
1  &  1  &  1  &  1  &  1  &   1  &  1 \\
6  &  4  &  2  &  0  & -2  &  -4  & -6 \\
15 &  5  & -1  & -3  & -1  &   5  & 15 \\
20 &  0  & -4  &  0  &  4  &   0  & -20 \\
15 & -5  & -1  &  3  & -1  &  -5  & 15 \\
6  & -4  &  2  &  0  & -2  &   4  & -6 \\
1  & -1  &  1  & -1  &  1  &  -1  &  1
\end{array}
\right].$
\end{center}
In these two matrices, one can verify the properties related to the
Krawtchouk matrix given in Proposition \ref{prop3a}.

To determine all the Walsh spectrum values of $f\in{\cal SB}_n$, it
is enough to multiply $((-1)^{f_0},\ldots,(-1)^{f_n})$ with the
$(n+1)\times(n+1)$ Krawtchouk matrix. Applying Krawtchouk matrix,
the analysis of the Walsh spectra of symmetric functions becomes
combinatorially interesting. Elements of a Krawtchouk matrix have
nice combinatorial properties and they follow nice
symmetry~\cite{IK96} too. We list some of them in the following
proposition.

\begin{proposition}
\label{prop3a}
1. $K_0(k, n)=1, K_1(k, n) = n-2k$,\\
2. $(i+1)K_{i+1}(k, n)=(n-2k)K_i(k,n)$\\
   $-(n-i+1)K_{i-1, n}(k, n)$,\\
3. $K_i(k, n)= (-1)^k K_{n-i}(k, n)$,\\
4. $\binom{n}{k} K_i(k, n)= \binom{n}{i} K_k(i, n)$,\\
5. $K_i(k, n) = (-1)^i K_i(n-k, n)$,\\
6. $(n-k)K_i(k+1, n) = (n-2i) K_i(k, n) - k K_i(k-1, n)$,\\
7. $(n-i+1)K_i(k,n+1)=(3n-2i-2k+1)K_i(k,n)$\\
   $-2(n-k)K_i(k,n-1)$.
\end{proposition}

\subsection{Implementation of Symmetric Boolean Functions}

The symmetric Boolean functions can be efficiently implemented. As
described in~\cite{sympoly}, the circuit complexity of $n$-variable
symmetric Boolean functions is $4.5n + o(n)$. It is known that given
a classical circuit $f$, there is a quantum circuit of comparable
efficiency which computes the transformation $U_f$ that takes input
like $|x, y\rangle$ and produces output like $|x, y \oplus
f(x)\rangle$. Thus, we will consider that for $f \in {\cal SB}_n$,
the quantum circuit $U_f$ can be efficiently implemented using
$O(n)$ circuit complexity.

\section{Algorithms}

\subsection{Find a Special Symmetric Boolean Function which maximizes the Walsh Spectrum}

Consider that we want to maximize the Walsh spectrum value
corresponding to weight $w$ points and naturally, from the property
of symmetric functions, all of them will be equal. Now we present an
important combinatorial result to show how to find such symmetric
Boolean functions.

\begin{theorem} \label{thm2}
Consider $f \in {\cal SB}_n$. The function $f$, represented as
$re_f$, for which the Walsh spectrum corresponding to the $w$ weight
points will be maximized, can be written as
$$
f_i =     \begin{cases}
                    0,          & $if$\,\, K_i(w, n) > 0 \\
                    1,          & $if$\,\, K_i(w, n) < 0 \\
                    0\, $or$\, 1,   & $if$\,\, K_i(w, n) = 0
            \end{cases}
$$
\end{theorem}
\pf We have $W_f(\omega) = \sum_{i = 0}^n (-1)^{f_i} K_i(k, n)$. One
may note that the maximum value of $\sum_{i = 0}^n (-1)^{f_i} K_i(k,
n)$ is $\sum_{i = 0}^n |K_i(k, n)|$. This is attained when we take
the function of the form as described in the theorem. \qed

\begin{example}
\label{ex1} As example, consider $n = 6, w = k = 2$. In the
corresponding column of the $(6+1) \times (6+1)$ matrix, we get the
values as $1, 2, -1, -4, -1, 2, 1$. Thus, we will consider the
function with $re_f$ as $[0, 0, 1, 1, 1, 0, 0]$. For such an $f \in
{\cal SB}_6$, the Walsh spectrum values at the points $\omega$, such
that $w = wt(\omega) = 2$, will be maximized, which is
$1+2+1+4+1+2+1 = 12$.
\end{example}

\subsection{Walsh Spectrum of the Special Symmetric Boolean function by Combinatorial Property of Krawtchouk Matrix}

Next we present certain results related to column sum of Krawtchouk
matrix.
\begin{lemma}\label{lem1}
$\sum_{i=0}^n |K_i(\lceil\frac{n}{2}\rceil, n)| = \sum_{i=0}^n
|K_i(\lfloor\frac{n}{2}\rfloor, n)| =2^{\lceil\frac{n}{2}\rceil}$.
\end{lemma}
\pf Let us first prove this for even $n$.

Following Proposition~\ref{prop3a}(2), we have
\begin{eqnarray*}
(i+1)K_{i+1}(k, n) & = & (n-2k) K_i(k, n)\\
                   &   & -(n-i+1)K_{i-1, n}(k, n).
\end{eqnarray*}
For $n$ even, and $k = \frac{n}{2}$, we get,
$$K_{i+1}(\frac{n}{2}, n) = -\frac{n-i+1}{i+1} K_{i-1, n}(\frac{n}{2}, n).$$
That is, the recurrence relation follows:
$$K_{i}(\frac{n}{2}, n) = -\frac{n-i+2}{i} K_{i-2, n}(\frac{n}{2}, n),$$ with
the initial conditions $K_0(\frac{n}{2}, n) = 1$ and
$K_1(\frac{n}{2}, n) = 0$ as available from
Proposition~\ref{prop3a}(1). Thus one may note that for odd $i$,
$K_i(\frac{n}{2}, n) = 0$. Further, using induction, for even $i$,
we get
$$|K_i(\frac{n}{2}, n)| = \binom{\frac{n}{2}}{\frac{i}{2}}.$$
Thus, $$\sum_{i=0}^n |K_i(\frac{n}{2}, n)|=
\sum_{l=0}^{\frac{n}{2}}\binom{\frac{n}{2}}{l},$$ putting $i = 2l$.
Hence,
$$\sum_{i=0}^n |K_i(\frac{n}{2}, n)| = 2^{\frac{n}{2}}.$$
Now let us prove this for odd $n$.

For $n$ odd and $k = \frac{n-1}{2}$, from
Proposition~\ref{prop3a}(2) we get
\begin{eqnarray*}
K_{i+1}(\frac{n-1}{2},n) & = & \frac{1}{i+1}K_i (\frac{n-1}{2},n)\\
 & & - \frac{n-i+1}{i+1}K_{i-1}(\frac{n-1}{2},n).
\end{eqnarray*}
That is, the recurrence relation is as follows:
\begin{eqnarray*}
K_{i}(\frac{n-1}{2},n) & = & \frac{1}{i}K_{i-1} (\frac{n-1}{2},n)\\
& &  - \frac{n-i+2}{i}K_{i-2}(\frac{n-1}{2},n).
\end{eqnarray*}
One can now show by induction that
$$K_{2i}(\frac{n-1}{2},n) = K_{2i+1}(\frac{n-1}{2},n),\ \forall i, \ 1 \leq i \leq \frac{n-1}{2}.$$ Using the above two identities and induction,
one can verify that $|K_{2l}(\frac{n-1}{2},n)| =
\binom{\frac{n-1}{2}}{l}$. Thus, $$\sum_{i=0}^n
|K_i(\lfloor\frac{n}{2}\rfloor, n)| = 2\sum_{l=0}^{\frac{n-1}{2}}
\binom{\frac{n-1}{2}}{l},$$ where $i = 2l$. Hence, we get,
$$\sum_{i=0}^n |K_i(\lfloor\frac{n}{2}\rfloor, n)| = 2 \cdot 2^{\frac{n-1}{2}} = 2^{\frac{n+1}{2}}.$$

Using Proposition~\ref{prop3a}(5), we get that
$$\sum_{i=0}^n |K_i(\lfloor\frac{n}{2}\rfloor, n)| = \sum_{i=0}^n |K_i(\lceil\frac{n}{2}\rceil, n)|.$$ That completes the proof. \qed

\begin{theorem}\label{thm3}
Let $f \in {\cal SB}_n$ be as explained in Theorem~\ref{thm2}
towards maximizing the Walsh spectrum values at weight
$\lceil\frac{n}{2}\rceil$ or $\lfloor\frac{n}{2}\rfloor$. Then,
$$\binom{n}{\lceil\frac{n}{2}\rceil}
(rw_f(\lceil\frac{n}{2}\rceil))^2 =
\binom{n}{\lfloor\frac{n}{2}\rfloor}
(rw_f(\lfloor\frac{n}{2}\rfloor))^2 \mbox{ is }
\Omega(\frac{2^{2n}}{\sqrt{n}}).$$
\end{theorem}
\pf The Walsh spectrum in this case is
\begin{eqnarray*}
rw_f(\lceil\frac{n}{2}\rceil) & = & \sum_{i=0}^n |K_i(\lceil\frac{n}{2}\rceil, n)|\\
= rw_f(\lfloor\frac{n}{2}\rfloor) & = & \sum_{i=0}^n
|K_i(\lfloor\frac{n}{2}\rfloor, n)|.
\end{eqnarray*}
Thus the total sum of the squares of the Walsh spectrum values at
weight $\lceil\frac{n}{2}\rceil$ or $\lfloor\frac{n}{2}\rfloor$ is
$$\binom{n}{\lceil\frac{n}{2}\rceil}(\sum_{i=0}^n |K_i(\lceil\frac{n}{2}\rceil, n)|)^2$$
which is $\Omega(\frac{2^{2n}}{\sqrt{n}})$, by Stirling's
approximation. \qed

One may similarly note that for the trivial cases of $w = 0$ or $n$,
if one chooses $f \in {\cal SB}_n$ following Theorem~\ref{thm2},
then $\binom{n}{w} (rw_f(w))^2 = 2^{2n}$. However, proving the
result similar to Theorem~\ref{thm3} for any $n$ and any weight $w$,
in general, seems to be quite tedious. Thus we make detailed
enumerations to obtain $c(n) = \min_{w = 0}^n \frac{\binom{n}{w}
\left(rw_f(w)\right)^2}{\frac{2^{2n}}{\sqrt{n}}}$ that has been
verified for $n \leq 1000$ and we note that the values stabilize as
$c(999) = 1.24793$ and $c(1000) = 0.797685$. The graph of this is
plotted in Figure~\ref{dickplot} for $1 \leq n \leq 100$, the points
for odd $n$ are coming above and those for even $n$ are coming
below. Since we are not providing a proof of this, we refer this as
follows.

\begin{figure}[t]
\centering
\includegraphics[width=0.49\textwidth]{dick.eps}
\caption{Plot of $c(n)$ vs $n$ for $1 \leq n \leq 100$.}
\label{dickplot}
\end{figure}

\begin{fact}
\label{fact1} Let $f \in {\cal SB}_n$ be as explained in
Theorem~\ref{thm2} towards maximizing the Walsh spectrum values at
weight $w$. Then the total sum of the squares of the Walsh spectrum
values at weight $w$, $\binom{n}{w} \left(rw_f(w)\right)^2$, is
$\Omega(\frac{2^{2n}}{\sqrt{n}})$.
\end{fact}
The proof of the fact seems to be quite tedious and elusive and we
leave it as an open problem.

\subsection{Relation between Deutsch-Jozsa algorithm and the Walsh Spectrum of Symmetric Boolean Function}

Given $f$ is either constant or balanced, if the corresponding
quantum implementation $U_f$ is available,
Deutsch-Jozsa~\cite{qDJ92} provided a quantum algorithm that decide
in constant time which one it is. Let us now describe our
interpretation of Deutsch-Jozsa algorithm in terms of Walsh spectrum
values. We denote the operator for Deutsch-Jozsa algorithm as
\begin{equation}
{\cal D}_f = H^{\otimes n} U_f H^{\otimes n}, \nonumber
\end{equation}
where the Boolean function $f$ is available as an oracle $U_f$. For
brevity, we abuse the notation and do not write the auxiliary qubit,
i.e., $\frac{|0\rangle - |1\rangle}{\sqrt{2}}$ and the corresponding
output in this case.

Now one can observe that~\cite{maitra}
\begin{eqnarray}\nonumber
{\cal D}_f |0\rangle^{\otimes n}    &   =   &   \sum_{z \in \{0,
1\}^n}\sum_{x \in \{0, 1\}^n} \frac{(-1)^{x \cdot z \oplus
f(x)}}{2^n}|z\rangle \\ \nonumber
                                    &   =   &   \sum_{z \in \{0, 1\}^n} \frac{W_f(z)}{2^n} |z\rangle. \nonumber
\end{eqnarray}
Note that the associated probability with a state $|z\rangle$ is
$\frac{W_f^2(z)}{2^{2n}}$. Hence we have the following technical
result as pointed out in~\cite{maitra} with our interpretation for
symmetric functions.

\begin{proposition}
\label{prop1} Given an $n$-variable Boolean function $f$, ${\cal
D}_f |0\rangle^{\otimes n}$ produces a superposition of all states
$z \in \{0, 1\}^n$ with the amplitude $\frac{W_f(z)}{2^n}$
corresponding to each state $|z\rangle$. Specially, if $f \in {\cal
SB}_n$, then the amplitude corresponding to $|z\rangle$ is
$\frac{\sum_{i = 0}^n (-1)^{f_i} K_i(wt(z), n)}{2^n}$.
\end{proposition}

\subsection{Algorithm 1: Deutsch-Jozsa Algorithm with Special Symmetric Function}
\label{dj}

Based on the overall properties, we propose a quantum algorithm as shown in the following algorithm.\\

\noindent {\bf Algorithm 1}\\
\begin{tabular}{p{240pt}} \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
1. Choose $f \in {\cal SB}_n$ as explained in Theorem~\ref{thm2} to maximize the Walsh spectrum values at weight $w$. \\
2. Use the Deutsch-Jozsa algorithm to obtain a symmetric $n$-qubit state $|S^n\rangle = \sum\limits_{x \in \{0, 1\}^n} a_{wt(x)} |x\rangle$, such that $\binom{n}{w} |a_w|^2$ is $\Omega(\frac{1}{\sqrt{n}})$.\\
3. Apply the parity measurement strategy. If the ancilla state is
measured at the basis $U^w|\zeta\rangle$, then $|D_w^n\rangle$ is
successfully obtained. Else go to Step 2 and iterate.\\   \hline
\end{tabular}\\ \\

The following result provides the estimate of complexity of our
algorithm.
\begin{theorem}
\label{thm5} Let $f \in {\cal SB}_n$ be as explained in
Theorem~\ref{thm2} towards maximizing the Walsh spectrum values at
weight $w$. Given Fact~\ref{fact1}, Deutsch-Jozsa algorithm produces
a symmetric $n$-qubit state (before the measurement) $|S^n\rangle =
\sum\limits_{x \in \{0, 1\}^n} a_{wt(x)} |x\rangle$, such that
$\binom{n}{w} |a_w|^2$ is $\Omega(\frac{1}{\sqrt{n}})$.
\end{theorem}
\pf The proof follows from Theorem~\ref{thm2}, Fact~\ref{fact1} and
Proposition~\ref{prop1}. \qed

Now the final step is to measure the symmetric state until we get
the target Dicke state by using parity measurement method
~\cite[Section IIIA]{d2}. Now we explain how to exploit the parity
measurement in our purpose. Note \cite[Section IIIA]{d2} assumes to
use $n$ dimensional qudit ancilla, but we consider a qudit
$|\zeta\rangle$ of dimension $n+1$ here. A certain unitary operator
$U$ is designed such that, $$|\zeta\rangle, U|\zeta\rangle,
U^2|\zeta\rangle, \ldots, U^{n-1}|\zeta\rangle, U^{n}|\zeta\rangle$$
are all orthogonal to each other and $U^{n+1}|\zeta\rangle =
|\zeta\rangle$. Since $|\zeta\rangle$ is an $(n+1)$-dimensional
state, one can indeed obtain a set of such $n+1$ orthogonal states.
The parity measurement is done on the
$$\{|\zeta\rangle, U|\zeta\rangle, U^2|\zeta\rangle, \ldots, U^{n}|\zeta\rangle\}$$ basis. Here $|\zeta\rangle$ is used as the target state. For the $n$-qubit control state $|x\rangle$, if it has weight $w$ then its corresponding target state, after application of this circuit, will become $U^{w}|\zeta\rangle$ (see Figure~\ref{fig1}). Now consider an $n$-qubit symmetric state as the control input, which is $|\tau\rangle =  \sum\limits_{x \in \{0, 1\}^n} a_w |x\rangle,$ where $w = wt(x)$, $\sum_{i = 0}^n \binom{n}{i} |a_i|^2 = 1$. After applying this circuit, one obtains $\sum\limits_{x \in \{0, 1\}^n} a_w |x\rangle U^w|\zeta\rangle$. Thus, if one measures in $$\{|\zeta\rangle, U|\zeta\rangle, U^2|\zeta\rangle, \ldots, U^{n}|\zeta\rangle\}$$ basis, then the state $|D_w^n\rangle$ will be obtained when the measurement result of the ancilla state is $U^w|\zeta\rangle$.

\begin{figure}[t]
\begin{center}
\begin{tikzpicture}[scale=0.82]
\node [inner sep=0pt,outer sep=0pt](p1) at (2,2.5) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (7,2.5) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (2,3.5) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (7,3.5) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (2,4) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (7,4) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (3,4) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (3,2) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (4,3.5) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (4,2) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (6,2.5) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (6,2) {}; \draw [-] (p1) -- (p2);
\node [inner sep=0pt,outer sep=0pt](p1) at (2,1.75) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (2.75,1.75) {}; \draw [thick] (p1) --
(p2); \node [inner sep=0pt,outer sep=0pt](p1) at (3.25,1.75) {};
\node [inner sep=0pt,outer sep=0pt](p2) at (3.75,1.75) {}; \draw
[thick] (p1) -- (p2); \node [inner sep=0pt,outer sep=0pt](p1) at
(4.25,1.75) {}; \node [inner sep=0pt,outer sep=0pt](p2) at
(5.75,1.75) {}; \draw [thick,dotted] (p1) -- (p2); \node [inner
sep=0pt,outer sep=0pt](p1) at (6.25,1.75) {}; \node [inner
sep=0pt,outer sep=0pt](p2) at (7,1.75) {}; \draw [thick] (p1) --
(p2); \node[rectangle, draw=black, minimum height=2, minimum
width=2] (H) at (3,1.75) {$U$}; \node[rectangle, draw=black, minimum
height=2, minimum width=2] (H) at (4,1.75) {$U$}; \node[rectangle,
draw=black, minimum height=2, minimum width=2] (H) at (6,1.75)
{$U$}; \fill[black] (3, 4) circle (0.1cm); \fill[black] (4, 3.5)
circle (0.1cm); \fill[black] (6, 2.5) circle (0.1cm); \node [inner
sep=0pt,outer sep=0pt] (t2) at (1.8,4) {$x_1$}; \node [inner
sep=0pt,outer sep=0pt] (t2) at (1.8,3.5) {$x_2$}; \node [inner
sep=0pt,outer sep=0pt] (t2) at (1.8,3) {\Large $\vdots$}; \node
[inner sep=0pt,outer sep=0pt] (t2) at (1.8,2.5) {$x_n$}; \node
[inner sep=0pt,outer sep=0pt] (t2) at (1.7,1.75) {$|\zeta\rangle$};
\node [inner sep=0pt,outer sep=0pt] (t2) at (7.75,1.75)
{$U^w|\zeta\rangle$};
\end{tikzpicture}
\caption{Generalized Parity Module as in~\cite[Fig. 1]{d2}. If
$wt(x) = w$, then the ancilla will become $U^w|\zeta\rangle$.}
\label{fig1}
\end{center}
\end{figure}

Since the probability of target Dicke state is
$\Omega(\frac{1}{\sqrt{n}})$, we should repeat the whole procedure
at most $O(\sqrt{n})$.

\begin{example}
\label{ex2} Let us have an example taking $n = 6, w = 2$ to outline
our method. In this case the Dicke state will be
$$|D^6_2\rangle = \sum\limits_{x \in \{0, 1\}^6, wt(x) = 2}
\frac{1}{\sqrt{15}} |x\rangle.$$ We start with is an $n = 6$
variable symmetric Boolean function having Walsh spectrum value at
each of the weight $w = 2$ point as $12$ (following
Theorem~\ref{thm2}, one may refer to Example~\ref{ex1} also). There
are $\binom{n}{w} = \binom{6}{2} = 15$ such points. The amplitude
associated with each point after the DJ algorithm is $\frac{12}{2^6}
= \frac{3}{16}$. Thus we get $\sum_{x \in \{0, 1\}^6, wt(x) = 2}
\frac{3}{16} |x\rangle + \sum_{x \in \{0, 1\}^6, wt(x) \neq 2} a_x
|x\rangle$ initially. Thus, the probability associated with
$|D^6_2\rangle$ will be $\binom{6}{2} (\frac{3}{16})^2 =
\frac{135}{256} = 0.52734375$ and hence one may note that the
probability that, after parity measurement, it will land into
$|D^6_2\rangle$ is quite high.
\end{example}

\subsection{Comparison with a Previous Method}
It was explained in~\cite{d2} how one can obtain $|D^n_w\rangle$
from $\frac{1}{2^{\frac{n}{2}}} \sum_{x \in \{0, 1\}^n} |x\rangle$.
However, the idea explained in~\cite[Section IIIA]{d2} works
efficiently only for $w = \lceil\frac{n}{2}\rceil$ or
$\lfloor\frac{n}{2}\rfloor$. The most general work in this direction
has appeared in~\cite{dickold} where biased Hadamard transformation
was exploited. The strategy of~\cite{dickold} uses biased Hadamard
transformation
$$\begin{pmatrix}
\sqrt{1-\frac{w}{n}} & \sqrt{\frac{w}{n}} \\
\sqrt{\frac{w}{n}} & -\sqrt{1-\frac{w}{n}}
\end{pmatrix}^{\otimes n}$$ on $|0\rangle^{\otimes n}$ such that the probability associated with $|D^n_w\rangle$ will be
$\binom{n}{w} (\frac{w}{n})^w (1-\frac{w}{n})^{n-w} \geq
\sqrt{\frac{2}{n\pi}}$, i.e., $\Omega(\frac{1}{\sqrt{n}})$. Thus,
the probability of our case and also in~\cite{dickold} are of the
same order. While the theoretical comparison of the exact
probability values seems elusive, we have made detailed enumerations
to observe that the exact probability values in our case are better
than that of~\cite{dickold}. Below we present enumeration results
towards that.

First we present two graphs to show the probability values
associated to $|D^n_w\rangle$ for all $w$, when $n = 999$ (to
represent odd case) or $1000$ (to represent even case). For our
case, it is $\binom{n}{w} \frac{(\sum_{i = 0}^n |K_i(w,
n)|)^2}{2^{2n}}$ (after application of Deutsch-Jozsa algorithm
without measurement), and for the case of \cite{dickold} it is
$\binom{n}{w} (\frac{w}{n})^w (1-\frac{w}{n})^{n-w}$ (after
application of biased Hadamard transform). From Figure~\ref{comp1a}
and Figure~\ref{comp1}, it is clear to note that our method provides
higher probability (the upper curve) in all the cases except $w = 0,
n$ (which are trivial ones) and $w = \frac{n}{2}$ for $n = 1000$.

\begin{figure}[t]
\centering
\includegraphics[width=0.49\textwidth]{odd.eps}
\caption{Plot of probabilities associated with $|D^n_w\rangle$
against $w$ in our case (above) and in~\cite{dickold} (below) for $n
= 999$.} \label{comp1a}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[width=0.49\textwidth]{compd1.eps}
\caption{Plot of probabilities associated with $|D^n_w\rangle$
against $w$ in our case (above) and in~\cite{dickold} (below) for $n
= 1000$.} \label{comp1}
\end{figure}
\begin{figure}[t]
\centering
\includegraphics[width=0.49\textwidth]{compd2.eps}
\caption{Plot of probabilities associated with $|D^n_w\rangle$
against $n$ in our case (above) and in~\cite{dickold} (below) for $n
= 4$ to $1000$ and $w = \lfloor \frac{n}{4}\rfloor$.} \label{comp2}
\end{figure}

In both figures, the present algorithm shows some variation of
probability when the weight is around $\lfloor \frac{n}{4} \rfloor$.
To check whether or not these cases still shows the higher
probability than the previous method, we look into the $w = \lfloor
\frac{n}{4} \rfloor$ case a little bit more. From
Figure~\ref{comp2}, one may note that our probability values (the
upper curve) are higher than the case of~\cite{dickold}. These
results explains the advantage of the use of a suitable symmetric
Boolean function which shows higher Walsh spectrum values for the
given weight.

\subsection{Algorithm 2: Additional Improvement by exploiting biased Hadamard Operator}

We have provided numerical evidences that using proper symmetric
Boolean functions in Deutsch-Jozsa algorithm provides better
probability than the use of biased Hadamard transform as described
in~\cite{dickold}. However, motivated by~\cite{dickold}, a natural
extension should be to couple biased Hadamard transform in
Deutsch-Jozsa algorithm instead of (unbiased) Hadamard transform.
Thus, let us refer to the general description of a Hadamard type
transformation (biased or unbiased) that can be written as $B_{r, n}
=
\begin{pmatrix}
\sqrt{1-\frac{r}{n}} & \sqrt{\frac{r}{n}} \\
\sqrt{\frac{r}{n}} & -\sqrt{1-\frac{r}{n}}
\end{pmatrix}$.
We will replace the standard notation of $w$ here by $r$ as we will
not restrict ourselves to integer values $w \in [0, \ldots, n]$, but
use any real number $r \in [0, n]$ to obtain the optimum probability
of success to get a Dicke state.

Instead of using the operator ${\cal D}_f = H^{\otimes n} U_f
H^{\otimes n}$, let us first describe the most general operator of
the form
\begin{equation}
{\cal D'}_f = B_{r_1, n}^{\otimes n} U_f B_{r_2, n}^{\otimes n},
\end{equation}
where $r_1, r_2$ are real numbers in $[0, n]$.

First, we consider the case when $r_1 = \frac{n}{2}$, i.e., $B_{r_1,
n} = H$, but $r_2 = r$ varies towards optimization. That is
\begin{equation}
{\cal D'}_f = H^{\otimes n} U_f B_{r, n}^{\otimes n}.
\end{equation}

One may note that the application of ${\cal D'}_f$ on
$|0\rangle^{\otimes n}$ will produce

${\cal D'}_f |0\rangle^{\otimes n} = \frac{1}{\sqrt{2^n}} \sum_{z
\in \{0, 1\}^n}$

$\left(\sum_{x \in \{0, 1\}^n} (-1)^{x\cdot z\oplus f(x)}
(1-\frac{r}{n})^{\frac{n-d(x,z)}{2}}
(\frac{r}{n})^{\frac{d(x,z)}{2}}\right) |z\rangle$, where $d(x, z)$
is the (Hamming) distance between two same length binary strings $x$
and $z$. Before proceeding further, we also have the following
technical result.
\begin{proposition}
Let ${\cal D'}_f = H^{\otimes n} U_f B_{r, n}^{\otimes n}$. If $f
\in {\cal SB}_n$ then ${\cal D'}_f |0\rangle^{\otimes n}$ is a
symmetric state.
\end{proposition}
\pf We need to prove that $\sum_{x \in \{0, 1\}^n} (-1)^{x\cdot
z\oplus f(x)} (1-\frac{r}{n})^{\frac{n-d(x,z)}{2}}
(\frac{r}{n})^{\frac{d(x,z)}{2}}$ is same for all the $z \in \{0,
1\}^n$ having the same Hamming weight. Let us consider $u, v \in
\{0, 1\}^n$ such that $u \neq v$, but $wt(u) = wt(v)$. Then we need
to prove that $\sum_{x \in \{0, 1\}^n} (-1)^{x\cdot u\oplus f(x)}
(1-\frac{r}{n})^{\frac{n-d(x,u)}{2}}
(\frac{r}{n})^{\frac{d(x,u)}{2}}$ $=$ $\sum_{x \in \{0, 1\}^n}
(-1)^{x\cdot v\oplus f(x)} (1-\frac{r}{n})^{\frac{n-d(x,v)}{2}}
(\frac{r}{n})^{\frac{d(x,v)}{2}}$.

The proof follows from the fact that $\sum_{x \in \{0, 1\}^n, wt(x)
= w} (-1)^{x\cdot u\oplus f(x)} (1-\frac{r}{n})^{\frac{n-d(x,u)}{2}}
(\frac{r}{n})^{\frac{d(x,u)}{2}} =$ $\sum_{x \in \{0, 1\}^n, wt(x) =
w} (-1)^{x\cdot v\oplus f(x)} (1-\frac{r}{n})^{\frac{n-d(x,v)}{2}}
(\frac{r}{n})^{\frac{d(x,v)}{2}}$, given that $f$ is symmetric. \qed

The main problem in this case is that we need to go for trial and
error by modifying the symmetric Boolean functions and trying out
different values of $\frac{r}{n}$. So far, we could not obtain the
exact characterization of symmetric functions while biased Hadamard
transform is used.
Based on this, we propose Algorithm 2 as follows.\\

\noindent {\bf Algorithm 2}\\
\begin{tabular}{p{240pt}} \hline
  % after \\: \hline or \cline{col1-col2} \cline{col3-col4} ...
1. Apply ${\cal D'}_f$ to $|0\rangle^{\otimes n}$ to obtain a
symmetric $n$-qubit state. The value of $\frac{r}{n}$ and the choice
of the symmetric Boolean
functions are achieved heuristically.\\
